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Algebra · real student question

Solve the inequality (x + 1)(x - 3) > 0.

Question

Solve the inequality

(x+1)(x−3)>0(x+1)(x-3)>0

Step-by-step solution

  1. Keep it factored. A product is positive when both factors share the same sign and negative when they differ. Expanding to x2−2x−3x^2-2x-3 would throw away exactly the information that makes this easy.

  2. Find the zeros of each factor.

    x+1=0⇒x=−1,x−3=0⇒x=3x+1=0\Rightarrow x=-1,\qquad x-3=0\Rightarrow x=3

    The product can only change sign at these two points, so they cut the line into (−∞,−1)(-\infty,-1), (−1,3)(-1,3) and (3,∞)(3,\infty).

  3. Test one value in each interval. At x=−2x=-2: (−1)(−5)=5>0(-1)(-5)=5>0 (both negative). At x=0x=0: (1)(−3)=−3<0(1)(-3)=-3<0 (signs differ). At x=4x=4: (5)(1)=5>0(5)(1)=5>0 (both positive). The pattern is positive, negative, positive.

  4. Confirm the pattern from the shape of the graph. Expanded, the leading coefficient is +1+1, so the parabola opens upward: it dips below the axis between its roots and rises above outside them. That is exactly the sign pattern found by testing — a useful independent check.

  5. Select the intervals and exclude the roots. The inequality is strict, so the points where the product is exactly 00 do not count:

    x<−1orx>3,(−∞,−1)∪(3,∞)x<-1\qquad\text{or}\qquad x>3,\qquad(-\infty,-1)\cup(3,\infty)

  6. Verify by scanning. Evaluating the product at 20012001 points from −10-10 to 1010 and comparing with the claimed solution set gives agreement at every single point ✓.

Answer

x<−1 or x>3,(−∞,−1)∪(3,∞)x<-1\ \text{or}\ x>3,\qquad(-\infty,-1)\cup(3,\infty)

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